Dev and Lap Splice Lengths - ETN-D-1-15 PDF
Dev and Lap Splice Lengths - ETN-D-1-15 PDF
Dev and Lap Splice Lengths - ETN-D-1-15 PDF
Technical Note
ETN-D-1-15
(f) Type and location of end-bearing splices; Since the 1971 Code, major changes were
made to the provisions for calculating ℓd in ACI
(g) Type and location of welded splices and oth- 318-89 and -95 [1989, 1995]. No major techni-
er required welding of reinforcing bars. cal revisions were introduced in the 1999 edition
through the current 2014 edition, i.e., the provi-
This Technical Note focuses on Items “(b)” sions for calculating ℓd in the 2014 Code are es-
and "(d)", i.e., on determining tension develop- sentially the same as those in the 1995, 1999,
ment lengths and tension lap splice lengths of re- 2002, 2005, 2008, and 2011 Codes. This Techni-
inforcing bars. A reinforcing bar must be “embed- cal Note discusses the provisions in ACI 318-14.
ded” or “anchored” a sufficient distance or length Several examples are presented to demonstrate
in concrete so the bar is capable of developing application of the two procedures for calculating
its design strength. The basic premise is the “an- ℓd .
chorage length” or “embedment length” must be
equal to or greater than the required tension de- 2014 ACI Building Code
velopment length of the bar given by the Code.
Under ACI 318-14, as with the 1995, 1999,
Regarding the other items, provisions in other
2002, 2005, 2008, and 2011 Codes, the Architect
parts of the Code include performance require-
/ Engineer has a choice of two procedures for cal-
ments for mechanical and welded splices. Fur-
culating ℓd , which are presented in Code Sections
ther information on mechanical splices is pre-
25.4.2.2 and 25.4.2.3.
sented in the CRSI publication, Reinforcing Bars:
Anchorages and Splices. Commentary Section Section 25.4.2.2. This section provides a
R26.6.4 discusses welded splices. The ACI Code short-cut approach for calculating ℓd . The expres-
cites “Structural Welding Code – Reinforcing sions for calculating ℓd are reproduced in Table 1,
Steel (AWS D1.4/D1.4M:2011)” as the standard which is based on Table 25.4.2.2 of the Code. Use
for welding reinforcing bars. of Section 25.4.2.2 requires selection of the appli-
cable expression from the four expressions given
Development Length. The concept of “de- in Table 1. The applicable expression is based on:
velopment length” of reinforcing bars was intro- • Bar size; expressions are given for #3
duced in the 318-71 ACI Building Code [1971]. through #6 bars, and for #7 bars and larger.
Table 1 – Tension Development Length – Section 25.4.2.2 in ACI 318-14*
Conditions Bar Sizes #3 to #6 Bar Sizes #7 to #18
Clear spacing of bars or wires being developed or
lap spliced not less than db , clear cover not less
than db , and stirrups or ties throughout ℓd not less
than the Code minimum; or
Clear spacing of bars or wires being developed or
( )
fy ψt ψe
db
25λ√fc´
(a) ( )
fy ψt ψe
db
20λ√fc´
(b)
lap spliced not less than 2db and concrete cover not
less than db
Other cases
( )
3fy ψt ψe
db
50λ√fc´
(c) ( )
3fy ψt ψe
db
40λ√fc´
(d)
= 1.0 for “other” bars
Section 25.4.2.3. This section presents a general ap-
The product of ψt ψe need not be taken greater than 1.7.
proach in which particular values of concrete cover and
bar spacing, as well as the amount of transverse rein-
Based on experience in fielding inquiries from design-
forcement, is taken into account. Code Eq. 25.4.2.3a in
ers and in presenting seminars, there seems to be a
Section 25.4.2.3 includes the effects of several of the
tendency among some Code users to categorize Sec-
major variables:
( ))
tion 25.4.2.3 as being applicable only to structural
3 fy ψ ψ ψ members with transverse reinforcement. Or that Sec-
t e s
ℓd = db (Code Eq. 25.4.2.3a) tion 25.4.2.3 is most advantageous for use with struc-
40 λ√f
c
´
( cb+Ktr
db
tural members having stirrups or ties. Presumably,
the presence of the K tr term in the denominator of
Eq. 25.4.2.3a has an influence for such actions.
The confinement term (cb + Ktr )/db is limited to a maxi-
mum value of 2.5.
The Code is clear as to the use or applicability of the
Atr = total cross-sectional area of all transverse rein- K tr term. At the end of Section 25.4.2.3, following the
forcement within the spacing s that crosses the equation for K tr , the Code states:
potential plane of splitting through the bars being
developed, in.2 “It shall be permitted to use K tr = 0 as a design sim-
cb = see discussion in text, in. plification even if transverse reinforcement is present.”
db = nominal diameter of bar, in.
Thus, for those structural members without trans-
f c´ = specified compressive strength of concrete, psi
verse reinforcement, or if the stirrups in beams or the
fy = specified yield strength of reinforcing bars, psi ties in columns are ignored, the part of the denominator
/
K tr = 40 Atr sn, in. of Eq.25.4.2.3a with the K tr term reduces to determining
the value of (cb /db) for the particular conditions. The value
n = number of bars being developed or lap spliced
along the plane of splitting “cb” is the smaller of: (1) one-half of the center-to-center
spacing of the bars; and (2) the distance from center of the
s = center-to-center spacing of transverse reinforce-
bar to the nearest concrete surface. The definition of “cb”
ment within ℓd , in.
presents new concepts. Center-to-center bar spacing (ac-
λ = 1.0 for normal weight concrete tually one-half of the c.–c. spacing) is used rather than the
= 0.75 for lightweight concrete clear spacing, which is used in Section 25.4.2.2. Instead
ψ = 1.0 for uncoated and galvanized bars of concrete cover to the bar as used in Section 25.4.2.2
e
= 1.5 for epoxy-coated or zinc and epoxy dual and prescribed in Section 20.6, cover as used in Section
coated bars with clear cover < 3db, 25.4.2.3 is the distance from the center of the bar to the
or clear spacing < 6db nearest concrete surface.
2 Tension Development and Lap Splice Lengths of Reinforcing Bars under ACI 318-14 [ETN-D-1-15]
Table 2 – Tension Lap Splice, ℓst
Max. percent of
As,provided/As,required As spliced within
required lap length
Splice
Type
ℓst
Over Length of splice
50 Class A Greater of: 1.0 ℓd and 12 in.
≥ 2.0
100 Class B
Greater of: 1.3 ℓd and 12 in.
< 2.0 All cases Class C
Section 25.5.1. This section presents the Code require- Concrete cover = 2.0 in. or 2.7db
ments for tension lap splices, ℓst. Table 2, which is based From Table 1; under the heading “Conditions” with
on Table 25.5.2.1 in the Code, defines the conditions un- clear spacing > 2db , concrete cover > db , and bar size
der which lap Class A or B can be used, where Class A is #6, the applicable expression is:
( )
defined as 1.00 ld, and Class Bis defined as 1.3 ld. fy ψt ψe
ℓd = db
Note that the 12 in. minimum is not imposed on ℓd if 25 λ√fc´
it is used to determine lap lengths, so the Code imposes
this limit after lap length has been calculated. For exam- For this example, the factors ψt , ψe and λ are equal
ple, if ℓd without the 12 in. minimum is determined to be to 1.0. Thus,
10 in., a Class A ℓst would be 1.0 x 10 = 10 in., increased (60,000)(1.0)(1.0)(0.75)
to 12 in. and Class B ℓst would be 1.3 x 10 = 13 in. Other ℓd =
25 (1.0)√ 4,000
key points for lap splices:
= 28.5 or 29 in.*
• Tension lap splices are not permitted for #14 or #18
bars. If the bars are epoxy-coated, the coating factor, ψe ,
• Lap spliced reinforcing bars are permitted not to be has to be determined. Because the concrete cover val-
in contact, but the maximum center-to-center spac- ue of 2.7db is less than 3db , the coating factor ψe = 1.5.
ing of lap splices bars cannot be the lesser of one- Thus, for the #6 epoxy-coated bars:
fifth the required lap splice length and 6 in. ℓd = 1.5(28.5) = 42.7 or 43 in.
• When calculated for lap splice lengths, ℓd cannot
be reduced by ratio of As,provided /As,required. (B) ℓd by Section 25.4.2.3
(
t e
cb+Ktr
s
)) db
Example No. 1 db
Given: An 8-in. thick slab is reinforced with #6 Grade
60 uncoated bars with a center-to-center spacing of For this solution, the factor ψs = 0.8 for the #6 bars,
10 in. Concrete cover is 2 in.; normal-weight concrete and the factors ψt , ψe and λ are equal to 1.0. Thus,
with fc´= 4,000 psi.
= 28.5 or 29 in.
Comments: The results are summarized in Table 3. Note
that ℓd for the uncoated #6 bars under Section 25.4.2.2 Using Section 25.4.2.3 and Code Eq. 25.4.2.3a:
is 71% longer than the ℓd required by Section 25.4.2.3.
For epoxy-coated #6 bars, Section 25.4.2.2 requires an cb is smaller of (0.75 + 0.75/2) = 1.1 in. √ governs
ℓd which is 65% longer than the ℓd required by Section or 10/2 = 5.0 in.
25.4.2.3. Similar results were observed for Class A and cb = 1.1 in.
Class B ℓst . (cb+ Kt r )/db = (1.1 + 0)/0.75 = 1.5 < 2.5, use 1.5
(
e
40 λ√fc´ cb+Ktr))
s
db
2014 Code
Section Uncoated
Class B ℓst
Epoxy-Coated
ℓd = (
3 60,000
40 (1.0)√ 4,000
(1.0)1.0(0.8)
1.5
)0.75
4 Tension Development and Lap Splice Lengths of Reinforcing Bars under ACI 318-14 [ETN-D-1-15]
Find: Check the required tension development length of
Calculate ℓd using Code Eq. 25.4.2.3a:
( ))
the #10 bars versus the available embedment.
Solution: 3 fy ψ ψ ψ
t e s
ℓd = db
40 λ√f
c
´
( cb+Ktr
db
ℓd =
3
( 60,000
40 (1.0)√ 3,000
(1.0)1.0(1.0)
2.5
1.27 )
= 41.7 or 42 in.
(
ℓd =
20 λ√fc´
)
fy ψt ψe
db
stirrups. Grade 60, uncoated bottom bars in the interior
span of a continuous beam. Other data are: bw = 24 in.;
h = 30 in.; concrete cover to the stirrups is 1.5 in.; normal-
For this example, the factors ψt , ψe and λ are equal to weight concrete with fc´ = 4,000 psi; #4 U-stirrups are
1.0. Thus, spaced at 13 in. on center and provided throughout ℓd .
(60,000)(1.0)(1.0)(0.75)
ℓd =
20 (1.0)√ 3,000
= 69.6 or 70 in.
If K tr is taken as zero:
(cb + K tr )/db = (2.25 + 0)/1.27
= 1.8 < 2.5, use 1.8
Figure 3 – Assumed Location of #10 Bar at Then ℓd = (45.2)(2.0)/1.8 = 50.2 or 51 in.
Corner of #4 Stirrup
This example shows a reduction in ℓd using Section
25.4.2.3 instead of Section 25.4.2.2 of 25% when taking
Bar spacing and concrete cover: the #4 stirrups into account, and 16% when the stirrups
From side face of beam to center of outermost #10 are neglected. The results are summarized in Table 5.
bar, the distance is: 1.5 (cover) + 0.5 (stirrup diameter)
+ 1.0 (X) = 3.0 in. Table 5 – Results of Example No. 3
c.–c. spacing of the 5 – #10 bars 2014 Code Section ℓd
= (24 – (2)(3))/4 12.2.2 61 in.
= 4.5 in. 25.4.2.3 (with K tr = 0.25) 46 in.
Clear spacing = 4.5 – 1.27 = 3.2 in. or 2.5db 25.4.2.3 (with K tr = 0) 51 in.
Concrete cover = 1.5 + 0.5 = 2.0 in. or 1.6db
Example No. 4
(A) ℓd by Section 25.4.2.2 Given: Consider the base slab of a cantilever retaining
wall. The concrete is normal weight with fc´ = 3,000 psi.
The applicable expression from Table 1 is: Assume that the #11 bars, spaced at 8 in. c. to c., are
( fy ψt ψe
)
required to resist the factored moment at Point A, i.e., the
ℓd = db tension ℓd cannot be reduced by the ratio of As (required)
20 λ√fc´
to As (provided).
For this example, the factors ψt , ψe and λ are equal to
Figure 4 – Base Slab of Retaining Wall
1.0. Thus,
(60,000)(1.0)(1.0)(1.27)
ℓd =
20 (1.0)√ 4,000
= 60.2 or 61 in.
( ))
Clear spacing of the bars = 8.0 – 1.41
3 fy ψ ψ ψ = 6.59 in. or 4.7db
t e s
ℓd = db
40 λ√f Concrete cover = 2 in. or 1.4db
c
´
( cb+Ktr
db The applicable expression from Table 1 is:
6 Tension Development and Lap Splice Lengths of Reinforcing Bars under ACI 318-14 [ETN-D-1-15]
Table 7 – Tension Development and Lap Splice Lengths for Bars in Walls, Slabs and Footings (ACI 25.4.2.3)
f’c = 3,000 psi
Concrete Cover = 0.75 in. Concrete Cover = 1.50 in. Concrete Cover = 2.00 in. Concrete Cover = 3.00 in.
Bar Lap
Uncoated Epoxy-Coated Uncoated Epoxy-Coated Uncoated Epoxy-Coated Uncoated Epoxy-Coated
Size Class
Top Other Top Other Top Other Top Other Top Other Top Other Top Other Top Other
A 13 12 17 15 13 12 17 15 13 12 17 15 13 12 17 15
#3
B 17 13 22 20 17 13 22 20 17 13 22 20 17 13 22 20
A 22 17 28 25 17 13 23 20 17 13 23 20 17 13 23 20
#4
B 28 22 37 32 23 17 29 26 23 17 29 26 23 17 29 26
A 32 24 41 37 22 17 28 25 22 17 28 25 22 17 28 25
#5
B 41 32 54 47 28 22 37 32 28 22 37 32 28 22 37 32
A 43 33 56 50 26 20 34 30 26 20 34 30 26 20 34 30
#6
B 56 43 73 64 34 26 44 39 34 26 44 39 34 26 44 39
A 69 53 90 80 43 33 55 49 38 29 49 43 38 29 49 43
#7
B 90 69 117 104 55 43 72 64 49 38 64 56 49 38 64 56
A 86 66 112 99 54 41 70 62 43 33 56 50 43 33 56 50
#8
B 111 86 146 128 70 54 91 80 56 43 73 64 56 43 73 64
A 104 80 136 120 66 51 86 76 53 41 70 61 48 37 63 56
#9
B 135 104 176 155 86 66 112 99 69 53 90 80 63 48 82 73
A 125 96 163 144 81 62 106 93 66 51 86 76 55 42 71 63
#10
B 162 125 212 187 105 81 137 121 85 66 111 98 71 55 93 82
A 146 113 191 169 97 74 126 111 79 61 103 91 61 47 79 70
#11
B 190 146 248 219 125 97 164 145 102 79 134 118 79 61 103 91
Table – Tension Development and Lap Splice Lengths for Bars in Walls, Slabs and Footings (ACI 25.4.2.3)
f’c = 4,000 psi
Concrete Cover = 0.75 in. Concrete Cover = 1.50 in. Concrete Cover = 2.00 in. Concrete Cover = 3.00 in.
Bar Lap
Uncoated Epoxy-Coated Uncoated Epoxy-Coated Uncoated Epoxy-Coated Uncoated Epoxy-Coated
Size Class
Top Other Top Other Top Other Top Other Top Other Top Other Top Other Top Other
A 12 12 15 13 12 12 15 13 12 12 15 13 12 12 15 13
#3
B 15 12 19 17 15 12 19 17 15 12 19 17 15 12 19 17
A 19 15 24 22 15 12 20 17 15 12 20 17 15 12 20 17
#4
B 24 19 32 28 20 15 25 22 20 15 25 22 20 15 25 22
A 28 21 36 32 19 15 24 22 19 15 24 22 19 15 24 22
#5
B 36 28 47 41 24 19 32 28 24 19 32 28 24 19 32 28
A 37 29 49 43 22 17 29 26 22 17 29 26 22 17 29 26
#6
B 48 37 63 56 29 22 38 34 29 22 38 34 29 22 38 34
A 60 46 78 69 37 28 48 42 33 25 43 38 33 25 43 38
#7
B 78 60 102 90 48 37 62 55 42 33 55 49 42 33 55 49
A 74 57 97 86 47 36 61 54 37 29 49 43 37 29 49 43
#8
B 96 74 126 111 60 47 79 70 48 37 63 56 48 37 63 56
A 90 69 117 104 57 44 75 66 46 36 60 53 42 32 55 48
#9
B 117 90 153 135 74 57 97 86 60 46 78 69 55 42 71 63
A 108 83 141 125 70 54 92 81 57 44 74 66 47 36 62 55
#10
B 140 108 183 162 91 70 119 105 74 57 97 85 61 47 80 71
A 127 98 166 146 84 64 109 97 68 53 89 79 52 40 69 60
#11
B 165 127 215 190 109 84 142 125 89 68 116 102 68 52 89 79
Notes:
1. Tabulated values are based on a minimum yield strength of 60,000 psi and normal-weight concrete. Lengths are in inches.
2. Tension development lengths and tension lap splice lengths are calculated per ACI 318-14, Sections 25.4.2.3 and 25.5.1, respectively, with bar
sizes limited to #3 through #11.
3. When the variable “cb” from ACI 25.4.2.3 was calculated, it was assumed that concrete cover controlled. That is, c.– c. spacing was assumed
to be greater than 1.0 db plus twice the concrete cover.
4. Lap splice lengths (minimum of 12 inches) are multiples of tension development lengths; Class A = 1.0 ℓd and Class B = 1.3 ℓd
(ACI 318 25.5.1). When determining the lap splice length, ℓd is calculated without the 12-inch minimum of ACI 25.4.2.1.
5. Top bars are horizontal bars with more than 12 inches of concrete cast below the bars.
6. For epoxy-coated bars, if the c.-c. spacing is at least 7.0 db and the concrete cover is at least 3.0 db, then lengths may be multiplied by 0.918
(for top bars) or 0.8 (for other bars).
7. For Grade 75 reinforcing bars, multiply the tabulated values by 1.25. For Grade 80 reinforcing bars, multiply the tabulated values by 1.33.
8. For lightweight concrete, divide the tabulated values by 0.75.
For this example, the bar location factor ψt = 1.3 for The available embedment length to the left of Point
top bars, and the factors ψe and λ are equal to 1.0. Thus A is 6 ft.-9 in. or 81 in. Because the required ℓd = 101 in.
is greater than the available embedment length, the #11
60,000)(1.3)(1.0)(1.41) bars cannot be anchored as straight bars according to
ℓd =
20 (1.0)√ 3,000 Section 12.2.2.
= 100.4 or 101 in.
ℓd = ( 3 fy
40 λ√f
c
´
(
ψ ψ ψ
t e
cb+Ktr
s
)) db For the second part of Example 1, the concrete cover
is only 0.75 in. From Table 8 for Lap Class A and “other”
db bars:
ℓd = 29 in. for uncoated bars
For this example, the bar location factor ψt = 1.3, and
the factors ψe , ψs and λ are equal to 1.0. Thus, Example No. 2T For the spread footing with uncoat-
ed #10 bars and concrete cover of 3 in. to the layer of
ℓd =
3
( 60,000
40 (1.0)√ 3,000
(1.3)1.0(1.0)
1.9
1.41 ) bars nearest the bottom, normal-weight concrete with
fc´ = 3,000 psi...
= 79.3 or 80 in.
Enter Table 7; for Lap Class A and “other” bars:
Because ℓd = 80 in. does not exceed the available em- ℓd = 42 in. for uncoated bars
bedment length of 81 in., the #11 bars can be anchored
as straight bars. This example clearly demonstrates the Example No. 3T Tables 7 and 8 are not intended for
significant reduction in ℓd that is possible, under certain and consequently are not applicable for closely-spaced
conditions, by using Section 25.4.2.3 instead of Section bars in beams. For the beam in Example 3, the value of
25.4.2.2. The computed ℓd of 80 in. by Section 25.4.2.3 cb would be governed by one-half of the c.–c. spacing of
is 21% shorter than the 101 in. computed by Section the bars, i.e., 2.25 in., rather than by the concrete cover
25.4.2.2. The results are summarized in Table 6. plus one-half of a bar diameter, i.e., 2.6 in.
Table 6 – Results of Example No. 4 Example No. 4T For the base slab of the cantilever
retaining wall with uncoated #11 bars spaced at 8 in. c.–
2014 Code Available Properly c., concrete cover of 2 in., normal-weight concrete with
ℓd
Section Embedment Anchored? fc´ = 3,000 psi...
25.4.2.2 101 in. 81 in. No
25.4.2.3 80 in. 81 in. Yes Enter Table 7; for Lap Class A and “top” bars:
ℓd = 79 in. for uncoated bars
Summary
Tabular Values Based on Section 25.4.2.3
This Technical Note discusses the provisions in Sec-
Tables 7 and 8 give values of ℓd , based on Code Sec-
tions 25.4.2.2 and 25.4.2.3 of the 2014 ACI Building
tion 25.4.2.3 and Eq. 25.4.2.3a, for walls, slabs and foot-
Code for determining the tension development lengths,
ings. The values for “Lap Class A” are also the values of
ℓd , of reinforcing bars. Several examples are presented
ℓd , because the required lap length for a Class A tension
to complement the discussion. The examples serve to
lap splice is 1.0 ℓd .
identify some of the conditions and the structural mem-
bers for which the more rigorous provisions in Section
An important restriction on the use of Tables 7 and
25.4.2.3 can be used advantageously.
8 is described in Note 3, i.e., it is assumed that the
value “cb“ in the quantity, (cb + K tr )/db, in Code Eq.
25.4.2.3a is governed by concrete cover rather than
by one-half the center-to-center spacing of the bars.
8 Tension Development and Lap Splice Lengths of Reinforcing Bars under ACI 318-14 [ETN-D-1-15]
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Concrete (ACI 318-71), American Concrete Institute, 465 pp.
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[1989], Building Code Requirements for Reinforced American Welding Society [2011], Structural Welding
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American Concrete Institute, Detroit, Michigan, 353 pp. Welding Society, Miami, Florida, 72 pp.
American Concrete Institute – ACI Committee 318 Concrete Reinforcing Steel Institute [2008], Reinforc-
[1995], Building Code Requirements for Structural Con- ing Bars: Anchorages and Splices, 5th Edition, Concrete
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Contributors: Dr. David P. Gustafson, P.E., S.E. and Anthony L. Felder, P.E., with subsequent
contributions from Neal S. Anderson, P.E., S.E..
Reference: Concrete Reinforcing Steel Institute-CRSI [2015], “Tension Development and Lap
Splice Lengths of Reinforcing Bars Under ACI 318-14,” CRSI Technical Note ETN-D-1-15, Con- 933 North Plum Grove Rd.
crete Reinforcing Steel Institute, Schaumburg, Illinois, 8 pp. Schaumburg, IL 60173-4758
p. 847-517-1200 • f. 847-517-1206
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